This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and 
and 
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
Learn more:
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
H2(g) +C2H4(g)→C2H6(g)
H-H +H2C =CH2→H3C-Ch3
2C -H bonds and one C-C bond are formed while enthalpy change (dH) of the reaction,
H-H: 432kJ/mol
C=C: 614kJ/mol
C-C: 413 kJ/mol
C-C: 347 kJ/mol
dH is equal to sum of the energies released during the formation of new bonds or negative sign, and sum of energies required to break old bonds or positive sign.
The bond which breaks energy is positive.
432+614 =1046kJ/mol
Formation of bond energy is negative
2(413) + 347 = 1173 kJ/mol
dH reaction is -1173 + 1046 =-127kJ/mol
